# Inelastic Collision or Conversation of Energy?

1. Nov 7, 2011

### en bloc

1. The problem statement, all variables and given/known data
Leonard wants to know how fast he can throw a ball, so he hangs a 2.3 kg target on a rope from a tree. He picks up a 0.5 kg ball of clay and throws it horizontally against the target. The clay sticks to the target and the clay and the target swing up to a height of 1.5m above the initial position. How fast did Leonard throw the ball of clay?

2. Relevant equations
KE+PE = KE+PE

3. The attempt at a solution
I drew it out and thought this equation might work, but i don't think it's right for many reasons.

1/2*.5kg*v^(2)=2.8kg*9.8*1.5m

v=12.8 m/s

unrealistically high velocity.

There's no way for me to check this answer, so can someone tell me if this is right?

2. Nov 7, 2011

### PeterO

You are trying to relate the kinetic energy of the ball before the collision to the kinetic energy of the combination after collision. This collision is not elastic, so that is not correct. Any collision wher the masses stick together is inelastic.

3. Nov 7, 2011

### ehild

It is correct, and not unrealistic. Think: You throw a stone and at the same time your friend starts to run. Which is faster: the stone or the man? What is the speed a man can run?

ehild

Edit: Of course it is wrong, the energy is not conserved during the collision, only after.

Last edited: Nov 7, 2011
4. Nov 7, 2011

### en bloc

I'm a bit confused here. Did i get the right answer, but just with not the best approach?

And i now see how this speed isn't unrealistically high, thanks.

Last edited: Nov 7, 2011
5. Nov 7, 2011

### PeterO

The answer is incorrect.

if you relate KE after collision to height reached, it shows the 2.8 kg will be travelling at between 5 and 6 m/s

Conservation of momentum will show that the 0.5 kg mass must have been travelling at a little over 30 m/s.

That is not unrerasonable, as a good baseball pitcher can throw things at up to 40 m/s. I suppose that if you "friend" was boasting about the speed at which he can throw things it may be possible, as if he couldn't throw very fast, he wouldn't be boasting.

6. Nov 7, 2011

### ehild

PeterO: you are right.

ehild

7. Nov 7, 2011

### en bloc

That's a huge mistake on my part. The whole point of conservation of energy is that everything in the system is conserved, so i can't add new mass. Thank you!

KE+PE = KE+PE

1/2 * 2.8 kg * v^(2) = 2.8kg * g * 1.5m
v=5.42 m/s

mv+mv=(m+m)v

.5 kg * v + 0 = (2.8kg)5.42 m/s
v= 30.4 m/s

8. Nov 7, 2011

### ehild

It is all right now, but never write such things that "KE+PE= KE +PE" or mv+mv=(m+m)v. They do not mean anything but "everyting is identical with itself". Use different notations and/or iindices: KE1+PE1= KE2 +PE2, m1v+m2v2=(m1+m2)u.

ehild

9. Nov 7, 2011

### technician

PeterO is right. After the collision you equate the PE to the KE of the combined target + ball of clay (2.8kg).
This gives you the velocity AFTER the collision.
MOMENTUM (not KE) is conserved during the collision so you use momentum before = momentum after to calculate the velocity of the ball of clay.
This question is an example of something called a 'ballistic pendulum'.... used to determine the speed of bullets by firing into a massive block and noting how high it swings. At my school we use this technique to determine the speed of air rifle pellets!!!